Integrand size = 8, antiderivative size = 51 \[ \int \frac {\arcsin (a x)}{x} \, dx=-\frac {1}{2} i \arcsin (a x)^2+\arcsin (a x) \log \left (1-e^{2 i \arcsin (a x)}\right )-\frac {1}{2} i \operatorname {PolyLog}\left (2,e^{2 i \arcsin (a x)}\right ) \]
-1/2*I*arcsin(a*x)^2+arcsin(a*x)*ln(1-(I*a*x+(-a^2*x^2+1)^(1/2))^2)-1/2*I* polylog(2,(I*a*x+(-a^2*x^2+1)^(1/2))^2)
Time = 0.05 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.90 \[ \int \frac {\arcsin (a x)}{x} \, dx=\arcsin (a x) \log \left (1-e^{2 i \arcsin (a x)}\right )-\frac {1}{2} i \left (\arcsin (a x)^2+\operatorname {PolyLog}\left (2,e^{2 i \arcsin (a x)}\right )\right ) \]
ArcSin[a*x]*Log[1 - E^((2*I)*ArcSin[a*x])] - (I/2)*(ArcSin[a*x]^2 + PolyLo g[2, E^((2*I)*ArcSin[a*x])])
Time = 0.31 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.16, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {5136, 3042, 25, 4200, 25, 2620, 2715, 2838}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\arcsin (a x)}{x} \, dx\) |
\(\Big \downarrow \) 5136 |
\(\displaystyle \int \frac {\sqrt {1-a^2 x^2} \arcsin (a x)}{a x}d\arcsin (a x)\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\arcsin (a x) \tan \left (\arcsin (a x)+\frac {\pi }{2}\right )d\arcsin (a x)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \arcsin (a x) \tan \left (\arcsin (a x)+\frac {\pi }{2}\right )d\arcsin (a x)\) |
\(\Big \downarrow \) 4200 |
\(\displaystyle 2 i \int -\frac {e^{2 i \arcsin (a x)} \arcsin (a x)}{1-e^{2 i \arcsin (a x)}}d\arcsin (a x)-\frac {1}{2} i \arcsin (a x)^2\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 i \int \frac {e^{2 i \arcsin (a x)} \arcsin (a x)}{1-e^{2 i \arcsin (a x)}}d\arcsin (a x)-\frac {1}{2} i \arcsin (a x)^2\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle -2 i \left (\frac {1}{2} i \arcsin (a x) \log \left (1-e^{2 i \arcsin (a x)}\right )-\frac {1}{2} i \int \log \left (1-e^{2 i \arcsin (a x)}\right )d\arcsin (a x)\right )-\frac {1}{2} i \arcsin (a x)^2\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle -2 i \left (\frac {1}{2} i \arcsin (a x) \log \left (1-e^{2 i \arcsin (a x)}\right )-\frac {1}{4} \int e^{-2 i \arcsin (a x)} \log \left (1-e^{2 i \arcsin (a x)}\right )de^{2 i \arcsin (a x)}\right )-\frac {1}{2} i \arcsin (a x)^2\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle -2 i \left (\frac {1}{4} \operatorname {PolyLog}\left (2,e^{2 i \arcsin (a x)}\right )+\frac {1}{2} i \arcsin (a x) \log \left (1-e^{2 i \arcsin (a x)}\right )\right )-\frac {1}{2} i \arcsin (a x)^2\) |
(-1/2*I)*ArcSin[a*x]^2 - (2*I)*((I/2)*ArcSin[a*x]*Log[1 - E^((2*I)*ArcSin[ a*x])] + PolyLog[2, E^((2*I)*ArcSin[a*x])]/4)
3.1.6.3.1 Defintions of rubi rules used
Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ ((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp [((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[d*(m/(b*f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x )))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[1/(d*e*n*Log[F]) Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) ))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol ] :> Simp[I*((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I Int[(c + d*x)^ m*E^(2*I*k*Pi)*(E^(2*I*(e + f*x))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))), x] , x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[( a + b*x)^n*Cot[x], x], x, ArcSin[c*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]
Time = 0.12 (sec) , antiderivative size = 111, normalized size of antiderivative = 2.18
method | result | size |
derivativedivides | \(-\frac {i \arcsin \left (a x \right )^{2}}{2}+\arcsin \left (a x \right ) \ln \left (1-i a x -\sqrt {-a^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, i a x +\sqrt {-a^{2} x^{2}+1}\right )+\arcsin \left (a x \right ) \ln \left (1+i a x +\sqrt {-a^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, -i a x -\sqrt {-a^{2} x^{2}+1}\right )\) | \(111\) |
default | \(-\frac {i \arcsin \left (a x \right )^{2}}{2}+\arcsin \left (a x \right ) \ln \left (1-i a x -\sqrt {-a^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, i a x +\sqrt {-a^{2} x^{2}+1}\right )+\arcsin \left (a x \right ) \ln \left (1+i a x +\sqrt {-a^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, -i a x -\sqrt {-a^{2} x^{2}+1}\right )\) | \(111\) |
-1/2*I*arcsin(a*x)^2+arcsin(a*x)*ln(1-I*a*x-(-a^2*x^2+1)^(1/2))-I*polylog( 2,I*a*x+(-a^2*x^2+1)^(1/2))+arcsin(a*x)*ln(1+I*a*x+(-a^2*x^2+1)^(1/2))-I*p olylog(2,-I*a*x-(-a^2*x^2+1)^(1/2))
\[ \int \frac {\arcsin (a x)}{x} \, dx=\int { \frac {\arcsin \left (a x\right )}{x} \,d x } \]
\[ \int \frac {\arcsin (a x)}{x} \, dx=\int \frac {\operatorname {asin}{\left (a x \right )}}{x}\, dx \]
\[ \int \frac {\arcsin (a x)}{x} \, dx=\int { \frac {\arcsin \left (a x\right )}{x} \,d x } \]
\[ \int \frac {\arcsin (a x)}{x} \, dx=\int { \frac {\arcsin \left (a x\right )}{x} \,d x } \]
Time = 0.05 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.80 \[ \int \frac {\arcsin (a x)}{x} \, dx=\ln \left (1-{\mathrm {e}}^{\mathrm {asin}\left (a\,x\right )\,2{}\mathrm {i}}\right )\,\mathrm {asin}\left (a\,x\right )-\frac {\mathrm {polylog}\left (2,{\mathrm {e}}^{\mathrm {asin}\left (a\,x\right )\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}}{2}-\frac {{\mathrm {asin}\left (a\,x\right )}^2\,1{}\mathrm {i}}{2} \]